Too easy Answer =3 Draw a perpendicular inside the isosceles. This is its height, then use Pythagorean. Height = 6.93 Since the perpendicular line creates another triangle, use Pythagorean to calculate the other side. This = 4 but 4= x + 1/2 the length of the isosceles or 1 4-1 = x =3 Answer
Not sure why all this trig. Drop an altitude from the vertex in the isosceles triangle, it has length sqrt(48) and splits the base in half, then (x+1)^2+48=64 so x+1=4 and x=3. That's your second method, though.
put midpoint m for dc so that MD=MC=1, AMB and AMD are right triangles. For AMD, AM^2 = AD^2 - MD^2 (pythagoras) = 7^2 - 1^2 = 48 ---------------------1 For AMB, BM^2 = AB^2 - AM^2 BM^2 = 64 - 48 (by 1) = 16 BM = 4 (length>0) BD = BM-MD = 4-1 = 3 x = 3 EDIT: I commented this after the 1st method.. the second method is similar to mine.
This is the perfect case to apply Stewart's theorem. Since the cevian AD is just known --> X=3. For Stewart's theorem and its demonstration here is the link below en.wikipedia.org/wiki/Stewart's_theorem
Drop an perpendicular from A to BC it will bisect ADC into two right triangles with length (1;h;7) h^2=(7^2-1^2) h^2=49-1=48 Let’s look at the right triangle with sides (h;1+x;8) By Pythagoras 8^2= (1+x)^2+h^2 64= (1+x)^2+48 16 = (1+x)^2 Take the square root on both sides 4 = 1+x or -4=1+x 3=x or -5=x Since x is a length it’s bigger than zero So x=3 is the correct answer
Drop a perpendicular from A to DC at E. As CA = AD, ∆CAD is an isosceles triangle, so AE bisects ∆CAD and forms two congruent right triangles ∆AEC and ∆DEA. As DC = 2, DE = EC = DC/2 = 1. Triangle ∆DEA: DE² + AE² = AD² 1² + AE² = 7² AE² = 49 - 1 = 48 AE = √48 = 4√3 As ∠BEA = 90°, ∆BEA is a right triangle. Ad DE = 1 and BD = x, BE = x+1. Triangle ∆BEA: BE² + AE² = AB² (x+1)² + 48 = 8² x² + 2x + 1 + 48 - 64 = 0 x² + 2x - 15 = 0 (x+5)(x-3) = 0 x = -5 ❌ | x = 3
The answer is x=15. And I thought that BOTH methods showed that this problem was MUCH easier than it looks!!! I will have to compare that problem with other problems on your channel. The first methods is Law of Cosines applied twice and the other is using the Pythagorean Theorem twice because of added length. Differing lengths. I think that this problem seems familiar!!!
Too easy Answer =3
Draw a perpendicular inside the isosceles. This is its height, then use Pythagorean. Height = 6.93
Since the perpendicular line creates another triangle, use Pythagorean to calculate the other side. This = 4
but 4= x + 1/2 the length of the isosceles or 1
4-1 = x =3 Answer
Not sure why all this trig. Drop an altitude from the vertex in the isosceles triangle, it has length sqrt(48) and splits the base in half, then (x+1)^2+48=64 so x+1=4 and x=3. That's your second method, though.
Yep. This one can be worked out sans paper'n'pencil.
Simplest and better solution 👋
put midpoint m for dc so that MD=MC=1, AMB and AMD are right triangles.
For AMD,
AM^2 = AD^2 - MD^2 (pythagoras) = 7^2 - 1^2 = 48 ---------------------1
For AMB,
BM^2 = AB^2 - AM^2
BM^2 = 64 - 48 (by 1) = 16
BM = 4 (length>0)
BD = BM-MD = 4-1 = 3
x = 3
EDIT: I commented this after the 1st method.. the second method is similar to mine.
This is the perfect case to apply Stewart's theorem.
Since the cevian AD is just known --> X=3.
For Stewart's theorem and its demonstration here is the link below
en.wikipedia.org/wiki/Stewart's_theorem
x=3
Drop an perpendicular from A to BC it will bisect ADC into two right triangles with length (1;h;7)
h^2=(7^2-1^2)
h^2=49-1=48
Let’s look at the right triangle with sides (h;1+x;8)
By Pythagoras
8^2= (1+x)^2+h^2
64= (1+x)^2+48
16 = (1+x)^2
Take the square root on both sides
4 = 1+x or -4=1+x
3=x or -5=x
Since x is a length it’s bigger than zero
So x=3 is the correct answer
Drop a perpendicular from A to DC at E. As CA = AD, ∆CAD is an isosceles triangle, so AE bisects ∆CAD and forms two congruent right triangles ∆AEC and ∆DEA. As DC = 2, DE = EC = DC/2 = 1.
Triangle ∆DEA:
DE² + AE² = AD²
1² + AE² = 7²
AE² = 49 - 1 = 48
AE = √48 = 4√3
As ∠BEA = 90°, ∆BEA is a right triangle. Ad DE = 1 and BD = x, BE = x+1.
Triangle ∆BEA:
BE² + AE² = AB²
(x+1)² + 48 = 8²
x² + 2x + 1 + 48 - 64 = 0
x² + 2x - 15 = 0
(x+5)(x-3) = 0
x = -5 ❌ | x = 3
ADC=α cosα/2=√(8*1/2*7)=2/√7(formule di Briggs)..teorema del coseno 64=x^2+49-2*7*x*cos(180-α)=x^2+49+14xcosα=x^2+49+14x(1/7)...x^2+2x-15=0...x=3
13:24 This can also be done as
(x+1)^2=64-48=16=(+/-4)^2
Then
x+1=4 (or x+1=-4)
x=3 (or x=-5)
Here's another way to see this :
Since sin(B) = sqrt(48)/8 = sqrt(3)/2 we have that angle B is 60 degrees, so (x + 1)/8 = 1/2 from which x = 3.
*Teorema de Stewart:*
AC²BD+ AB²CD - AD²BC=BC BD DC
asnwer=3cm isit
h²=7²-(2/2)²=48 → (X+1)²+h²=8²→ X=3.
Gracias y un saludo cordial.
АD*2=(X/X+2)AC*2+(2/X+2)AB*2-2X , X=3 . (Stewarts theorem)
*Teorema de Stewart:*
AC²BD+ AB²CD - AD²BC=BC BD DC
Nice
X=3
The answer is x=15. And I thought that BOTH methods showed that this problem was MUCH easier than it looks!!! I will have to compare that problem with other problems on your channel. The first methods is Law of Cosines applied twice and the other is using the Pythagorean Theorem twice because of added length. Differing lengths. I think that this problem seems familiar!!!
If x is 15 then ABD is degenerate, seems unlikely without me even trying the problem first.
Always belittling the channel and making mistakes. Be more humble and seek to learn more!
I stand corrected. I meant x=3. I should have clarified that that the hypotenuse is 5. Silly me!